Simpson’s Rule

Simpson’s rule estimates a definite integral by fitting parabolas to the curve, two subintervals at a time. With an even number n of subintervals of width Δx, it adds the function values with weights 1, 4, 2, 4, …, 2, 4, 1 and multiplies by Δx/3. It is usually far more accurate than the trapezoidal rule.

Before this lesson: Riemann Sums, Trapezoidal Rule and Definite Integrals

Why parabolas work so well

The trapezoidal rule joins points with straight lines. A curve bends, so straight lines miss some area. Simpson’s rule joins each group of three points with a parabola instead, and a parabola can bend with the curve.

That is why n must be even. Each parabola covers two subintervals, so the subintervals have to come in pairs.

Simpson’s rule formula

\int_a^b f(x)\,dx \approx S_n = \dfrac{\Delta x}{3}\Big[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)\Big]

Here \Delta x = \dfrac{b - a}{n}, x_k = a + k\,\Delta x, and n is even. The two ends get weight 1. Inside, the weights alternate 4, 2, 4, 2, … and the last inside point gets 4.

1. Find Δx = (b − a)/n and list the x-values.
2. Find f at each x-value.
3. Multiply by the weights 1, 4, 2, 4, …, 4, 1 and add.
4. Multiply the sum by Δx/3.

Error bound: if |f^{(4)}(x)| \le K on [a, b], then the error is at most \dfrac{K(b - a)^5}{180n^4}. Doubling n cuts this bound by a factor of 16.

Simpson’s rule examples

Example 1: A parabola is exact

Use Simpson’s rule with n = 2 to estimate \int_0^2 x^2\,dx.

\Delta x = \dfrac{2 - 0}{2} = 1

x_0 = 0, \; x_1 = 1, \; x_2 = 2

f(0) = 0, \; f(1) = 1, \; f(2) = 4

0 + 4(1) + 4 = 8

S_2 = \dfrac{1}{3}(8) = \dfrac{8}{3}

\int_0^2 x^2\,dx = \dfrac{2^3}{3} - 0 = \dfrac{8}{3}

Answer: S_2 = \dfrac{8}{3}, which is exact. Simpson’s rule is exact for any polynomial of degree 3 or less.

Example 2: Estimating ln 3

Use Simpson’s rule with n = 4 to estimate \int_1^3 \dfrac{1}{x}\,dx. Compare it with the exact value, ln 3.

\Delta x = \dfrac{3 - 1}{4} = 0.5

x = 1, \; 1.5, \; 2, \; 2.5, \; 3

f = 1, \; \dfrac{2}{3}, \; \dfrac{1}{2}, \; \dfrac{2}{5}, \; \dfrac{1}{3}

1 + 4\left(\dfrac{2}{3}\right) + 2\left(\dfrac{1}{2}\right) + 4\left(\dfrac{2}{5}\right) + \dfrac{1}{3}

= 1 + \dfrac{8}{3} + 1 + \dfrac{8}{5} + \dfrac{1}{3} = 6.6

S_4 = \dfrac{0.5}{3}(6.6) = 1.1

\ln 3 \approx 1.098612

1.1 - 1.098612 = 0.001388

Answer: S_4 = 1.1, off by only about 0.0014.

Example 3: An integral with no formula

Estimate \int_0^1 e^{-x^2}\,dx with n = 4. This integral has no antiderivative made of basic functions, so a numerical method is the way to go.

\Delta x = \dfrac{1 - 0}{4} = 0.25

f(0) = 1, \; f(0.25) \approx 0.939413, \; f(0.5) \approx 0.778801

f(0.75) \approx 0.569783, \; f(1) \approx 0.367879

1 + 4(0.939413) + 2(0.778801) + 4(0.569783) + 0.367879

= 1 + 3.757652 + 1.557602 + 2.279132 + 0.367879 = 8.962265

S_4 = \dfrac{0.25}{3}(8.962265) \approx 0.746855

Answer: about 0.74686. The true value is 0.746824…, so the error is about 0.00003.

Example 4: Choosing n with the error bound

How large must n be so that Simpson’s rule estimates \int_1^3 \dfrac{1}{x}\,dx with error at most 0.0001?

f(x) = x^{-1}, \; f^{(4)}(x) = 24x^{-5}

K = 24(1)^{-5} = 24

\dfrac{24(3 - 1)^5}{180n^4} \le 0.0001

\dfrac{768}{180n^4} \le 0.0001

n^4 \ge \dfrac{768}{180(0.0001)} \approx 42{,}667

n \ge 14.37

n = 16

Answer: n = 16 subintervals is enough.

Example 5: Area of a pond

A surveyor measures the width of a pond every 10 meters along its 80-meter length. The widths, in meters, are 0, 18, 26, 30, 32, 29, 24, 15 and 0. Estimate the pond’s area.

\Delta x = 10, \; n = 8

0 + 4(18) + 2(26) + 4(30) + 2(32) + 4(29) + 2(24) + 4(15) + 0

= 0 + 72 + 52 + 120 + 64 + 116 + 48 + 60 + 0 = 532

S_8 = \dfrac{10}{3}(532) \approx 1773.3

Answer: the pond’s area is about 1773 square meters. Simpson’s rule works with table data, as long as the points are evenly spaced and n is even.

Common mistakes

Related methods

Simpson’s rule from the trapezoidal and midpoint rules

If T_n is the trapezoidal estimate and M_n is the midpoint estimate with n subintervals, then S_{2n} = \dfrac{T_n + 2M_n}{3}. Their errors tend to have opposite signs, and this average cancels most of them.

Comparing the methods

Here K₂ bounds |f″| and K₄ bounds |f⁽⁴⁾|. Because Simpson’s error shrinks like 1/n⁴, it usually needs far fewer points.

Practice problems

1. Use Simpson’s rule with n = 4 to estimate \int_0^4 x^3\,dx.
2. Use Simpson’s rule with n = 4 to estimate \int_0^{\pi} \sin x\,dx.
3. Use Simpson’s rule with n = 4 to estimate \int_0^1 \dfrac{1}{1 + x^2}\,dx. Multiply by 4 to estimate π.
4. For n = 6, how many function values do you need, and what are the weights?
5. Use Simpson’s rule with n = 2 to estimate \int_2^4 \ln x\,dx. Compare with the exact value, 6 ln 2 − 2.
6. A car’s speed, in meters per second, is 0, 12, 20, 24 and 26 at t = 0, 2, 4, 6 and 8 seconds. Estimate the distance it travels.
7. Find the error bound for Simpson’s rule with n = 4 on \int_0^2 e^x\,dx.
8. How large must n be for Simpson’s rule to estimate \int_0^1 e^x\,dx with error at most 0.000001?
9. Use Simpson’s rule with n = 2 on \int_{-1}^3 (2x^3 - x + 5)\,dx and explain why the answer is exact.
10. For \int_0^2 x^4\,dx, T₂ = 9 and M₂ = 5.125. Find S₄ from them, and compare with the exact value.

\Delta x = 1; \; f = 0, 1, 8, 27, 64

0 + 4(1) + 2(8) + 4(27) + 64 = 192

S_4 = \dfrac{1}{3}(192) = 64

\Delta x = \dfrac{\pi}{4}; \; f = 0, \tfrac{\sqrt{2}}{2}, 1, \tfrac{\sqrt{2}}{2}, 0

0 + 4\left(\tfrac{\sqrt{2}}{2}\right) + 2(1) + 4\left(\tfrac{\sqrt{2}}{2}\right) + 0 = 4\sqrt{2} + 2 \approx 7.656854

S_4 = \dfrac{\pi}{12}(4\sqrt{2} + 2) \approx 2.00456

\Delta x = 0.25; \; f = 1, \tfrac{16}{17}, 0.8, 0.64, 0.5

1 + 4\left(\tfrac{16}{17}\right) + 2(0.8) + 4(0.64) + 0.5 \approx 9.424706

S_4 = \dfrac{0.25}{3}(9.424706) \approx 0.785392

4(0.785392) \approx 3.14157

n + 1 = 7

1, 4, 2, 4, 2, 4, 1

\Delta x = 1; \; f \approx 0.693147, \; 1.098612, \; 1.386294

0.693147 + 4(1.098612) + 1.386294 = 6.473889

S_2 = \dfrac{1}{3}(6.473889) \approx 2.157963

6\ln 2 - 2 \approx 2.158883

\Delta t = 2, \; n = 4

0 + 4(12) + 2(20) + 4(24) + 26 = 210

S_4 = \dfrac{2}{3}(210) = 140

K = e^2 \approx 7.389056

\dfrac{e^2(2 - 0)^5}{180(4)^4} = \dfrac{32e^2}{46{,}080}

\approx 0.005131

K = e

\dfrac{e(1 - 0)^5}{180n^4} \le 0.000001

n^4 \ge \dfrac{e}{180(0.000001)} \approx 15{,}101.6

n \ge 11.09

\Delta x = 2; \; f(-1) = 4, \; f(1) = 6, \; f(3) = 56

S_2 = \dfrac{2}{3}\big(4 + 4(6) + 56\big) = \dfrac{2}{3}(84) = 56

\Big[\tfrac{x^4}{2} - \tfrac{x^2}{2} + 5x\Big]_{-1}^{3} = 51 - (-5) = 56

S_4 = \dfrac{T_2 + 2M_2}{3} = \dfrac{9 + 2(5.125)}{3}

= \dfrac{19.25}{3} \approx 6.41667

\int_0^2 x^4\,dx = \dfrac{2^5}{5} = 6.4

Simpson’s rule FAQ

What is Simpson’s rule?

Simpson’s rule is a way to estimate a definite integral. It fits parabolas through groups of three points and adds the areas under them, using weights 1, 4, 2, 4, …, 4, 1 and the factor Δx/3.

Why does n have to be even in Simpson’s rule?

Each parabola covers two subintervals. To cover the whole interval with parabolas, the subintervals must come in pairs, so n is even.

Is Simpson’s rule more accurate than the trapezoidal rule?

Usually, yes. Its error shrinks like 1/n⁴, while the trapezoidal rule’s error shrinks like 1/n². It is also exact for cubics.

What is the error bound for Simpson’s rule?

The error is at most K(b − a)⁵/(180n⁴), where K is the largest value of |f⁽⁴⁾(x)| on [a, b].

Summary

• Simpson’s rule: S_n = (Δx/3)[f(x₀) + 4f(x₁) + 2f(x₂) + … + 4f(xₙ₋₁) + f(xₙ)].
• n must be even; the weights are 1, 4, 2, 4, …, 2, 4, 1.
• It is exact for polynomials of degree 3 or less.
• Error bound: K(b − a)⁵/(180n⁴), with K bounding |f⁽⁴⁾|.
• It works on evenly spaced table data, too.

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